(B) The self-inductance $L$ of a solenoid is given by $L = \frac{\mu N^2 A}{l}$,where $\mu = \mu_0 \mu_r$.Initially,$L_1 = \frac{\mu_0 N^2 A}{l} = 0.1

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(B) The self-inductance $L$ of a solenoid is given by $L = \frac{\mu N^2 A}{l}$,where $\mu = \mu_0 \mu_r$.Initially,$L_1 = \frac{\mu_0 N^2 A}{l} = 0.18\, mH$.When a core of relative permeability $\mu_r = 900$ is inserted,the new self-inductance $L_2$ becomes $L_2 = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_r L_1$.Substituting the values,we get $L_2 = 900 \times 0.18\, mH = 162\, mH$.

Class 12 Physics Electromagnetic Induction Self Induction

Easy

The coefficient of self-inductance of a solenoid is $0.18\, mH$. If a core of soft iron of relative permeability $900$ is inserted,then the coefficient of self-inductance will become nearly.....$mH$.

A
$5.4$
B
$162$
C
$0.006$
D
$0.0002$

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Electromagnetic Induction

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Self Induction

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In a circular conducting coil,when current increases from $2 \,A$ to $18 \,A$ in $0.05 \,s$,the induced $e.m.f.$ is $20 \,V$. The self-inductance of the coil is.....$mH$.

Medium

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In the circuit shown below,a long time after the key $K$ is closed,the reading in the ammeter is $20 \,mA$. What was the reading (in $mA$) immediately after it was closed?

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$A$ graph of magnetic flux $(\phi)$ versus current $(I)$ is shown for $4$ different inductors $P, Q, R, S$. The minimum value of inductance is for inductor:

EasyMHT CET 2024

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The coefficient of self-inductance of a solenoid is $0.18\, mH$. If a core of soft iron of relative permeability $900$ is inserted,then the coefficient of self-inductance will become nearly.....$mH$.

Similar Questions

When the current through an inductor is changed from $2 \,A$ to $6 \,A$ in time $2 \,s$, the emf induced in it is $3 \,V$. Then the inductance of the inductor is (in $\,H$)

$A$ coil of $100$ turns carries a current of $5\, mA$ and creates a magnetic flux of $10^{-5} \, Wb$. The inductance is.....$mH$.

In a circular conducting coil,when current increases from $2 \,A$ to $18 \,A$ in $0.05 \,s$,the induced $e.m.f.$ is $20 \,V$. The self-inductance of the coil is.....$mH$.

In the circuit shown below,a long time after the key $K$ is closed,the reading in the ammeter is $20 \,mA$. What was the reading (in $mA$) immediately after it was closed?

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